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| Question 2887:  I need your help on these three problems....................
 (1).
 -4     5
 -5
 The determinant of the matrix A is 1. Find the missing element of the matrix A
 a.1
 b.3
 c.6
 d.–24
 (2). Find the determinant of the matrix C.
 C=
 4  0  –7
 3  0  6
 -5  2  3
 
 a.45
 b.24
 c.-90
 d.-21
 
 (3). Use Cramer’s Rule to solve y.
 x-4y=4
 4x-y=1
 a.-8/15
 b. –1
 c. 17/5
 d. 0
 
 Answer by xcentaur(357)
      (Show Source): 
You can put this solution on YOUR website! [1] -4     5
 -5     X
 "The determinant of the matrix A is 1."
 
 
 In matrix X=
 Determinant = ab - cd
 
 
 Determinant = (-4*X)-(-5*5)
 = -4X+25
 Since we know the determinant is one,
 1=-4X+25
 -4X=-24
 X=-24/-4
 =6
 Therefore,correct answer is: c.6
 
 [2]
 Matrix C=
 
 In matrix X of order 3=
 
 Determinant of X is=
 =x1(y2z3-y3z2)-x2(y1z3-y3z1)+x3(y1z2-z1y2)
 Determinant of C is=
 = 4[0*3-(2*6)]-0[3*3-(-5*6)]+(-7)[3*2-(-5*0)]
 = 4[0-12]-0[9+30]-7[6-0]
 = 4[-12]-0[39]-7[6]
 = -48-0-42
 = -90
 
 Therefore,correct answer is: c.-90
 
 [3]
 Cramer's Rule - Given the system
 p1x + q1y = c1 ... [eqn1]
 p2x + q2y = c2 ... [eqn2]
 Determinant detA = p1q2 - p2q1
 Determinant detX = c1q2 - c2q1
 Determinant detY = p1c2 - p2c1
 Then x=detX/detA
 Then y=detY/detA
 
 
 Cramer’s Rule to solve y.
 x-4y=4
 4x-y=1
 
 
 Consider the determinant formed by the x,y and constant term of the given equations.
 
 
 detA = p1q2 - p2q1
 =(1*-1)-(4*-4)
 =-1-(-16)
 =-1+16
 =15
 detY = p1c2 - p2c1
 =(1*1)-(4*4)
 =1-16
 =-15
 
 
 By Cramer's Rule,y=detY/detA
 y=-15/15
 =-1
 
 Therefore,the correct answer is: b. –1
 
 Hope this helps,
 best of luck.
 
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